2006 AMC 8 Problem 20

Attempt Problem 20 of the 2006 AMC 8 below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2006 AMC 8 solutions, or check the answer key.

All problems are used with official legal permission of the Mathematical Association of America (MAA).

20.

A singles tournament had six players. Each player played every other player only once, with no ties. If Helen won 44 games, Ines won 33 games, Janet won 22 games, Kendra won 22 games and Lara won 22 games, how many games did Monica win?
00
11
22
33
44
Answer: C
Concepts:combinations
Small hint:
Count how many games are played in a six-player round robin.
Big hint:
Every game has exactly one winner, so total wins equal total games.
Solution:
In every match, there was exactly one winner. There are 6⋅5÷2=156 \cdot 5 \div 2 = 15 games and therefore 1515 wins.
There are already 4+3+2+2+2=13 4 + 3 + 2 + 2 + 2 = 13 wins accounted for, so Monica won 15−13=215 - 13 = 2 games.
Thus, C is the correct answer.
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